find-n-1-sin-nx-n-

Question Number 28614 by abdo imad last updated on 27/Jan/18

f i n d \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} .

Commented by abdo imad last updated on 30/Jan/18

l e t d e v e l o p p t h e f u n c t i o n f (x) = \frac{x}{2} p e r i o d i c b y 2 π a n d o d d

f (x) = \sum_{n = 1}^{\infty} a_{n} s i n (n x) a n d a_{n} = \frac{2}{2 π} \int_{[T]} \frac{x}{2} s i n (n x) d x

= \frac{1}{π} \int_{0}^{π} x s i n (n x) d x a n d b y p a r t s

{π a_{n} = \frac{- x}{n} c o s (n x)]}_{0}^{π} - \int_{0}^{π} \frac{- 1}{n} c o s (n x) d x

= \frac{- π}{n} {(- 1)}^{n} + \frac{1}{n^{2}} {[s i n (n x)]}_{0}^{π} = \frac{π {(- 1)}^{n - 1}}{n} s o

a_{n} = \frac{{(- 1)}^{n - 1}}{n} a n d

\frac{x}{2} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} s i n (n x) l e t d o t h e c h . = π - t s o

\frac{π - t}{2} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} s i n (n π - n t) b u t

s i n (n π - n t) = s i n (n π) c o s (n t) - c o s (n π) s i n (n t)

= {(- 1)}^{n - 1} s i n (n t) \Rightarrow \frac{π - t}{2} = \sum_{n = 1}^{\infty} \frac{s i n (n t)}{n} f i n a l l y w e h a v e

\sum_{n = 1}^{\infty} \frac{s i n (n x)}{n} = \frac{π - x}{2} .

Commented by abdo imad last updated on 30/Jan/18

t h e c o n v e r g e n c e o f t h i s s e r i e i s a s s u r e d b y A b e l D i r i c h l e t

t h e o r e m l e t r e m e m b e r t h i s t h e o r e m . i f Σ a_{n} (x) v_{n} (x) i s a

s e r i e o f f u n c t i o n (a_{n} > 0 a n d v_{n} > 0) / a_{n} d e c r e a s e a n d a_{n} {(x)}_{n \to \infty} \to 0

a n d \exists m > 0 / ∣ \sum_{k = n_{0}}^{n} v_{k} (x) ∣⩽ m s o t h e s e r i e c o n v e r g e s .